MATHEMATICS COLLEGE

which statement correctly defines the interquartile range for the set of data represented by the box plot A 14-1=13 , B 11-6=5, c 11-1=10, d 6-1=5​

Answers

Answer 1
Answer:

Answer:

B

Step-by-step explanation:
Answer 2
Answer:

Answer:

The answer is B, 11-6=5

Step-by-step explanation.

Since the range of both quartiles is 11 and 6. You would subtract both of these to get the answer. Which in this case, is 5.

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Suppose we are interested in bidding on a piece of land and we know one other bidder is interested. The seller announced that the highest bid in excess of $10,000 will be accepted. Assume that the competitor's bid x is a is a random variable that is uniformly distributed between $10,000 and $15,000.a. Suppose you bid $12,000. What is the probability that your bid will be accepted? (please show calculations)
b. Suppose you bid $14,000. What is the probability that your bid will be accepted? (please show calculations)
c. What amount should you bid to maximize the probability that you get the property? (please show calculations)d. Suppose you know someone who is willing to pay you $16,000 for the property. Would you consider bidding less than the amount in part (c)? Why or why not?

Answers

Answer:

Step-by-step explanation:

(a)

The bid should be greater than $10,000 to get accepted by the seller. Let bid x be a continuous random variable that is uniformly distributed between

$10,000 and $15,000

The interval of the accepted bidding is [ {\rm{\$ 10,000 , \$ 15,000}], where b = $15000 and a = $10000.

The interval of the provided bidding is [$10,000,$12,000]. The probability is calculated as,

\begin{array}{c}\nP\left( {X{\rm{ < 12,000}}} \right){\rm{ = }}1 - P\left( {X > 12000} \right)\n\n = 1 - \int\limits_(12000)^(15000) {\frac{1}{{15000 - 10000}}} dx\n\n = 1 - \int\limits_(12000)^(15000) {\frac{1}{{5000}}} dx\n\n = 1 - \frac{1}{{5000}}\left[ x \right]_(12000)^(15000)\n\end{array}

=1- ([15000-12000])/(5000)\n\n=1-0.6\n\n=0.4

(b)  The interval of the accepted bidding is [$10,000,$15,000], where b = $15,000 and a =$10,000. The interval of the given bidding is [$10,000,$14,000].

\begin{array}{c}\nP\left( {X{\rm{ < 14,000}}} \right){\rm{ = }}1 - P\left( {X > 14000} \right)\n\n = 1 - \int\limits_(14000)^(15000) {\frac{1}{{15000 - 10000}}} dx\n\n = 1 - \int\limits_(14000)^(15000) {\frac{1}{{5000}}} dx\n\n = 1 - \frac{1}{{5000}}\left[ x \right]_(14000)^(15000)\n\end{array} P(X<14,000)=1-P(X>14000)

=1- ([15000-14000])/(5000)\n\n=1-0.2\n\n=0.8

(c)

The amount that the customer bid to maximize the probability that the customer is getting the property is calculated as,  

The interval of the accepted bidding is [$10,000,$15,000],

where b = $15,000 and a = $10,000. The interval of the given bidding is [$10,000,$15,000].

\begin{array}{c}\nf\left( {X = {\rm{15,000}}} \right){\rm{ = }}\frac{{{\rm{15000}} - {\rm{10000}}}}{{{\rm{15000}} - {\rm{10000}}}}\n\n{\rm{ = }}\frac{{{\rm{5000}}}}{{{\rm{5000}}}}\n\n{\rm{ = 1}}\n\end{array}

(d)  The amount that the customer bid to maximize the probability that the customer is getting the property is $15,000, set by the seller. Another customer is willing to buy the property at $16,000.The bidding less than $16,000 getting considered as the minimum amount to get the property is $10,000.

The bidding amount less than $16,000 considered by the customers as the minimum amount to get the property is $10,000, and greater than $16,000 will depend on how useful the property is for the customer.

Smplify the following algebraic expression: 6(2y + 8) - 2(3y - 2)

Answers

Answer:

6y +52

Step-by-step explanation:

6(2y + 8) - 2(3y - 2)

Distribute

12y + 48 - 6y +4

Combine like terms

6y +52

Two number cubes are rolled for two seperate events: Event A is the event that the sum of numbers on both cubes is less than 10. Event B is the event that the sum of numbers on both cubes is a multiple of 3. Complete the conditional-probability formula for event B givem thaf event A occurs first by writing A and B in the blanks:

Answers

Answer:

Probability of getting sum less than

10 =  \frac{30}{36}=  \frac{5}{6}

Probability of getting a multiple of 3= \frac{20}{36}= \frac{5}{9}  

(A) P(B|A)= \frac{P(B∩A)}{P(A)} = \frac{15/36}{30/36}= \frac{15}{30}=0.5  

(B)P(A|B)= \frac{P(A∩B)}{P(B)}= \frac{15/36}{5/9}=0.75  

(C) {A∩B} = {3, 6, 9, 12, 15, 18}

(D) {A} = {1, 2, 3, 4, 5 ,6, 7, 8, 9}

-Hope this helps-

Help please i need all

Answers

Answer:

What grade are you in?

Step-by-step explanation:

6. 16 divided by 4 is 4

Then your keep going like 8 is 25 divided by 5 which is 5.

9. 120 divided by 4 is 30

10. 36 divided by 6 is 6
If you use a calculator it will be easier

1/2(x-6.5) = 3.16 what is x?

Answers

Answer:

x =12.82

Step-by-step explanation:

1/2(x-6.5) = 3.16

Multiply each side by 2

1/2(x-6.5) *2 = 3.16 *2

(x-6.5) = 6.32

Add 6.5 to each side

(x-6.5+6.5) = 6.32+ 6.5

x =12.82

Answer: 12.82

Step-by-step explanation:

1/2(x-6.5)= 3.16

x=12.82

After complaining about the bonus plans, you wake upon Conglomo's private island, where you break big rocks
into smaller rocks for $10 a day. There are 999 other
employees on the island, 998 of whom get paid the same
way you do. The last employee is your overseer, who is
paid $10 million per year.
What is the mean and median income for
workers on Super Happy Fun Island?
Which method of describing central tendency
better represents this data? Why?

Answers

Answers:

mean annual income =  $13,646.35

median annual income = $3,650

The median is a better measure of center

=======================================================

Explanation:

If you earn $10 a day, and do so for 365 days, then you earn 365*10 = 3650 dollars per year.

If there are 999 employees earning this amount, then the amount earned is 999*3650 = 3,646,350

Add on the 10,000,000 to get

10,000,000+3,646,350 = 13,646,350

The total amount earned is $13,646,350

Divide this over the 1000 people (999 workers + 1 boss)

We get: (13,646,350)/(1,000) = 13,646.35

The mean is annual income is $13,646.35

--------------------------------

The median is the middle most value. If you list out the pay amounts of just the workers, you'll get the list:

{3650, 3650, 3650, ..., 3650}

We will have 999 copies of 3650 listed out. You don't have to list all 999 of them. Just a few is a good start. The three dots indicate the pattern goes on until we reach the 999th item.

If we then tack on the overseer's pay, then we have the list

{3650, 3650, 3650, ..., 3650, 10 million}

I'm using "million" instead of the digits to avoid using commas here.

The middle won't change due to one item being tacked onto the end. The middle is going to be 3650 either due to it being part of the set, or we find the midpoint between 3650 and 3650, which averages out to 3650.

The median income is $3,650

-----------------------------------

The median income is the better measure of center because it represents where the workers are instead of some distant "midpoint" between the workers and the boss. No person is making $13,646.35 a year. They are either making $3,650 or they are making $10,000,000. There's nothing in between. So it's better go lean toward the larger group when deciding where the center should go. Think of it like a balance beam or a see-saw.

As you can see the median is not affected by outliers. We can change the "ten million" to something like "a hundred million" and the median would still be the same. I recommend you compute the mean if the overseer earns 100 million dollars, and you'll find the mean will increase dramatically. However, the median will stay where it's at. The only time the median will change is if we introduce elements that are somewhere around (either higher or lower) than 3650, and these values are close to 3650. But this won't change the center very much compared to the drastic changes the mean undergoes due to such a large outlier.

The general rule of thumb is: a large outlier to the right pulls the mean to the right. An outlier to the left pulls the mean to the left. In this case, the mean is pulled to the right. The data is skewed to the right (or positively skewed).

So again, the conclusion is that the median is the better measure of center.
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